proof of invariance of dimension

An application of the invariance of dimension theorem shows that ${ℝ}^n$ is homeomorphic to ${ℝ}^m$ if and only if $m=n$. Already this is a difficult question. (We will assume $n\le m$ throughout this article.)

Simple arguments suffice for small dimensions.

• If $n=0$ cardinality is sufficient: there can be no bijection between ${ℝ}^0=\{0\}$ and ${ℝ}^n$, $m>0$, as the latter is uncountable.

• If $n=1$, then suppose $f:ℝ\to {ℝ}^m$ is a homeomorphism with $m>1$. Then certainly the following restriction of $f$

  $$f:ℝ-\{0\}\to {ℝ}^m-\{f(0)\}$$

  is also a homeomorphism. Yet, as $m>1$, ${ℝ}^m-\{f(0)\}$ is (path) connected but $ℝ-\{0\}$ is not connected. Thus this restriction of$f$ cannot be a homeomorphism so indeed the original $f$ could not be a homeomorphism.

Unfortunately neither of these two arguments extends well to the cases where $n,m>1$. Indeed even the case for $n=1$ requires a reasonable amount of work to fill in the details. However, the latter approach does provide the necessary hint for a full solution.

To solve the problem outright depends on algebraic invariants from homology, asurprisingly big hammer for such a basic topological question. But the conceptual steps are still basic, and we will attempt to highlight them in our exposition of the proof.

Let $U$ and $V$ be non-empty open subsets of ${ℝ}^n$ and ${ℝ}^m$ respectively. Assume that $f:U\to V$ is a homeomorphism.

Choose a point $x\in U$ (akin to the point we removed when $n=1$.) Then consider the relative homology groups $H_i(U,U-\{x\})$, $i\in \mathbb{N}$. As $U$ is open we may apply the Excision Theorem (axiom) to claim $H_i(U,U-\{x\})\cong H_i({ℝ}^n,{ℝ}^n-\{x\})$ – basically, to look at a punctured open disk it to look at a punctured ${ℝ}^n$. Now we look at the induced long exact sequence from the relative pair $({ℝ}^n,{ℝ}^n-\{x\})$ and find $H_i({ℝ}^n,{ℝ}^n-\{x\})$ is isomorphic to the reduced homology ${\tilde{H}}_i({ℝ}^n-\{x\})$. But ${ℝ}^n-\{x\}$ contracts to the sphere $S^{n-1}$ – and homologoy preserves homotopy type – so we now have $H_i(U,U-\{x\})\cong H_i(S^{n-1})$. (Puncture a disk, and it deflates to a sphere of lower dimension.)

Now it is an exercise in homology to prove that $H_i(S^{n-1})=0$ if $i\ne 0,n-1$ and $\mathbb{Z}$ otherwise. In particular we are using the fact that the invariance of dimension of spheres is (more) easily established by the homology groups.

We now repeat the process with $V$. If $U$ and $V$ are indeed homeomorphic, then this process will result in isomorphic homology groups for every $i\in \mathbb{N}$. In particular,

$$\mathbb{Z}\cong H_{m-1}(S^{m-1})\cong H_{m-1}(V,V-\{f(x)\})\cong H_{m-1}(U,U-\{x\})\cong H_{m-1}(S^{n-1}).$$

Thus either $m=1$ which implies $n=0,1$ as $n\le m$, or $m=n$. If $n=0$ we have already seen $m=n$. So the result stands for all $m,n$.

For a detailed accounting of this theorem together with the necessary lemmas refer to:

Allen Hatcher, Algebraic Topology, Cambridge University Press, Cambridge, 2002. Available on-line at: http://www.math.cornell.edu/ hatcher/AT/ATpage.htmlhttp://www.math.cornell.edu/ hatcher/AT/ATpage.html